Let u=u_1 i+u_2 j+u_3 k be a unit vector in R^3 and v=1 6 (i+j+2 … - Vidyadip

MCQ

Let $\hat{u}=u_1 \hat{i}+u_2 \hat{j}+u_3 \hat{k}$ be a unit vector in $\mathbb{R}^3$ and $\hat{v}=\frac{1}{\sqrt{6}}(\hat{i}+\hat{j}+2 \hat{k})$. Given that there exists is(are) correct?

($A$) There is exactly one choice for such $\vec{v}$

($B$) There are infinitely many choices for such $\vec{v}$

($C$) If $\hat{u}$ lies in the $x y$-plane then $\left|u_1\right|=\left|u_2\right|$

($D$) If $\hat{u}$ lies in the $x z$-plane then $2\left|u_1\right|=\left|u_3\right|$

A
$B,C,A$
B
$B,D$
C
$B,A$
✓
$B,C$

✓

Answer

Correct option: D.

$B,C$

d
The correct options are

$B$ There are infinitely many choices for such $\vec{v}$

$C$ If $a$ lies in the $x y$-plane then $\left|u_1\right|=\left|u_2\right|$

$a=u_1 \hat{i}+u_2 \hat{j}+u_3 \hat{k}$

$\hat{w}=\frac{1}{\sqrt{6}}(\hat{i}+\hat{j}+2 \hat{k})$

$|0 \times \vec{v}|=1$

$\hat{w} \cdot(a \times \vec{v})=1$

$\Rightarrow|\hat{w}||a \times \vec{v}| \cos \alpha=1$

$\Rightarrow \cos a=1 \Rightarrow a=0^{\circ}$

So, $\hat{w} \perp \hat{a}$ and $\hat{w} \perp \vec{v}$

$\therefore a$ and $\vec{v}$ lies in the same plane.

So there will be infinite number of such $\vec{v}$.

Option $(3):$

If a lies in the $x y$-plane,

$a=u_1 \hat{i}+u_2 \hat{j}$

$a \cdot \hat{w}=0 \quad[\text { From (i)] }$

$\Rightarrow u_1+u_2=0 \Rightarrow\left|u_1\right|=\left|u_2\right|$

Option $(4):$

If a lies in the $x z$-plane,

$a=u_1 \hat{i}+u_3 \hat{k}$

$a \cdot \hat{w}=0$

$\Rightarrow u_1+2 u_3=0 \Rightarrow 2\left|u_3\right|=\left|u_1\right|$

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